# Product of singular matrices = the zero matrix

1. May 16, 2010

### clope023

1. The problem statement, all variables and given/known data

A = floor(10*rand(6)) (6x6 matrix with random numbers)

B = A'(transpose)

A(:,6) = -sum(B(1:5,' (sum row 1st through 5th row entries and place in the 6th column and then transpose and take the negative)

x = ones(6,1) (vector with 6 entries all equal to 1)

Ax = 0, x=/=0 implies A is a singular matrix since it is in contradiction to the fact that a nonsingular matrix implies x=0 is a unique solution of the equation Ax=0

[1:6] (vector with 6 entries from 1 to 6 respectively)

B = x*[1:6] (matrix column of 1's, column of 2's, ..., column of 6's)

The product AB should be the zero matrix. Why?

2. Relevant equations

det(A) = 0 => singular matrix

3. The attempt at a solution

set O = zero matrix

therefore AB = O take determinant of both sides det(AB) = det(O) => det(A)det(B) = det(O)

0det(B) = 0, therefore 0 = 0, I also calculated the determinant of matrix B with matlab and it seems it was singular as well so more easily I can calculate (0)(0) = det(O) => 0 = 0

I also attempted the question without the use of the determinant

assuming B is nonsingular

AB = O, ABinv(B) = Oinv(B), AI = O, A = O -> contradicts the original statement that A was not the zero matrix orignially this also implies there does not exist an inverse of B and thus B is also singular

I'm not sure if this is the correct to infer that the product of these two matrices is necessarily the zero matrix without using determinants from here any help would be appreciated thank you

Last edited: May 16, 2010
2. May 16, 2010

### D H

Staff Emeritus
Why, exactly, is Ax equal to zero? What does this mean regarding the matrix product AB?

Note: I am not asking because I disagree. Those questions are big fat hints.

3. May 16, 2010

### clope023

Ax = 0 would be because x=0 but x = ones(6,1) so this would imply that A = 0 however A was already a matrix =/= 0 so how could that be the zero matrix?

4. May 16, 2010

### D H

Staff Emeritus
No. x is not zero, and neither is A.

5. May 16, 2010

### clope023

B is equal to the vector [1:6] call it d

therefore B = x*d

AB = A*x*d = 0*d = O

not sure if that's quite it but closer I think

6. May 16, 2010

### D H

Staff Emeritus
B is not equal to the vector 1:6.

Why is Ax zero? Yes, you were told that it is zero. Don't just take this as a given. Understanding exactly why Ax=0 will help you understand why AB=0.

7. May 16, 2010

### clope023

sorry B=x*[1:6] mistyped on the keyboard

well the multiplcation of A*x will equal the zero vector since the summation of each row multiplied by the column of 1's inside the vector x will be equal to zero therefore the product is equal to a vector of 0's, the same procedure will yield a matrix of zeros for the product AB? I think the explanation should be simpler than this no?

8. May 16, 2010

### D H

Staff Emeritus
Try again. Why exactly is Ax equal to zero? Write it out.

Hint: What is the sixth column of A?

9. May 16, 2010

### clope023

I'm really not sure what your getting at by emphasising 'exactly'

the 6th column of A is composed of the negative of the summation of entries in each column element's particular row

this is another reason why when multiplied by vector ones(1,6) the product Ax = 0, this implies many solutions?

I'm still not sure how it applies unless the multiplication of A*x is somehow transferred over into the multiplication of A*B

10. May 16, 2010

### D H

Staff Emeritus
That is *the* reason why Ax=0.

You appear to be thinking that because the determinant of some matrix is zero then that means that Ax will be zero for any vector x, and that AB will be zero for any matrix B. That is not the case. Example:

$$A=\bmatrix 1 & 0 \\ 0 & 0\endbmatrix$$

While this matrix A is not constructed along the lines of the problem at hand, it certainly is singular. Yet A2 is not the zero matrix. In fact, A2=A.

This matrix is analogous to the problem at hand:

$$A=\bmatrix 1 & -1 \\ 2 & -2\endbmatrix$$

This is once again singular, and

$$A\bmatrix 1 \\ 1 \endbmatrix = \bmatrix 0 \\ 0 \endbmatrix$$

However,

$$A\bmatrix 1 \\ 0 \endbmatrix = \bmatrix 1 \\ 2 \endbmatrix$$

Obviously not the zero vector.

Back to the problem at hand: Suppose you replaced the last one in the unitary x vector with a zero. In that case, Ax is almost certainly not equal to zero.

Last edited: May 16, 2010
11. May 16, 2010

### clope023

so it is simply that the multiplication of rows of A and columns new B matrix will add to the zero vector? I thought there'd be more to it than that

12. May 16, 2010

### D H

Staff Emeritus
THey won't add to the zero vector. AB is a matrix, not a vector.

13. May 16, 2010

### clope023

correction the multiplication adds up to zero elements in a matrix, correct? thank you very much for all the help and your patience btw

14. May 16, 2010

### D H

Staff Emeritus
Correct, and you're welcome.